For the PDE $u_{t} = u_{xx}$, the Crank-Nicolson Method is defined as :
\[(1+r)U^{n+1}_{m} - \frac{r}{2}(U^{n+1}_{m+1} + U^{n+1}_{m-1}) = (1-r)U^{n}_{m} + \frac{r}{2}(U^{n}_{m+1} + U^{n}_{m-1})\]
Where $r = k/h^{2}$.  To calculate the local truncation error, we set this equal to zero and replace the approximate values with exact values.%  Then we have
%\[(1+r)u(x_{m},t_{n+1}) - \frac{r}{2}(u(x_{m+1},t_{n+1}) + u(x_{m-1},t_{n+1})) - (1-r)u(x_{m},t_{n}) - \frac{r}{2}(u(x_{m+1},t_{n}) + u(x_{m-1},t_{n})) = 0\]
Then we take the Taylor Expansion, expanding about $u(x_{m},t_{n})$
\begin{eqnarray}
 & & (1+r)\left(u + ku_{t} + \frac{1}{2}k^{2}u_{tt} + \frac{1}{6}k^{3}u_{ttt}\right)  \\
 & & -\frac{r}{2}\left(u + hu_{x} + ku_{t} + \frac{1}{2}h^{2}u_{xx} + hku_{xt} + \frac{1}{2}k^{2}u_{tt} + \frac{1}{2}h^{2}ku_{xxt} + \frac{1}{2}hk^{2}u_{xtt}\right) \\
 & & -\frac{r}{2}\left(u - hu_{x} + ku_{t} + \frac{1}{2}h^{2}u_{xx} - hku_{xt} + \frac{1}{2}k^{2}u_{tt} + \frac{1}{2}h^{2}ku_{xxt} - \frac{1}{2}hk^{2}u_{xtt}\right) \\
 & & - (1-r)u \\
 & & - \frac{r}{2}\left(u + hu_{x} + \frac{1}{2}h^{2}u_{xx} + \frac{1}{6}h^{3}u_{xxx} \right) \\
 & & - \frac{r}{2}\left(u - hu_{x} + \frac{1}{2}h^{2}u_{xx} - \frac{1}{6}h^{3}u_{xxx} \right)
% & & u + hu_{x} + ku_{t} + \frac{1}{2}h^{2}u_{xx} + hku_{xt} + \frac{1}{2}k^{2}u_{tt} + \frac{1}{6}h^{3}u_{xxx} + \frac{1}{2}h^{2}ku_{xxt} + \frac{1}{2}hk^{2}u_{xtt} + \frac{1}{6}k^{3}u_{ttt}
\end{eqnarray}
If we cancel, and collect terms we have
\begin{eqnarray*}
& & u_{t}\left(k + rk - rk\right) \\
 & + & u_{tt}\left(\frac{1}{2}k^{2} + \frac{1}{2}rk^{2} - \frac{1}{2}rk^{2}\right) \\
 & + & u_{xx}\left(-\frac{1}{2}rh^{2} - \frac{1}{2}rh^{2}\right) \\
 & + & u_{xxt}\left(-\frac{1}{2}rh^{2}k\right) \\
 & + & u_{ttt}\left(\frac{1}{6}k^{3} + \frac{1}{6}rk^{3}\right)
\end{eqnarray*}
After applying the facts $u_{t} = u_{xx}$ and $r = k/h^{2}$, we are left with
\[u_{ttt}\left(\frac{1}{6}k^{3}(1 + \frac{k}{h^{2}})\right)\]
Notice that all the terms in (2),(3),(5),(6) canceled.  We need to look at more terms to get a handle on the error.  Since all terms in (2) and (3) with odd powers of $h$ will cancel, we will only look at those with even powers.  Then for the Taylor expansion we get
\begin{eqnarray}
 & & (1 + r)\frac{1}{6}k^{3}u_{ttt} \\
 & - & r\left(\frac{1}{6}k^{3}u_{ttt} + \frac{1}{24}h^{4}u_{xxxx} + \frac{1}{4}k^{2}h^{2}u_{xxtt} \right) \\ % + \frac{1}{24}k^{4}u_{tttt}\right) \\
 & - & r\frac{1}{24}h^{4}u_{xxxx}
\end{eqnarray}
Then we apply the identity $u_{ttt} = u_{xxtt}$ and $r = k/h^{2}$ to get
\[\frac{-1}{12}\left(k^{3}u_{ttt} + kh^{2}u_{xxxx}\right) + H.O.T.\]
Which gives us an LTE of $O(k^{3} + kh^{2})$.
